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For a conic section, two lines each of which passes through the intersection of the tangents to the conic at its points of intersection with the other line are called conjugate lines.

dude, is the answer correct?

nice sir... the above is the method i had in mind.

sandeep, last time he said its not open to everyone..so just guess it might be the same this time too.

Anyway, i wont let this weakness let me down again

For the case

,

and so on...

In general,

hence inequality simplifies to

i.e

for x > 1 which is obviosully true.

Similar method for x < 1 , with the differnce that

.

unfair!

u both didnt even wait for him to say if it was open to all!!

open to all?

diksha and punnima, there is a limit to how much patience we have...

For god's sake, stop posting rubbish. If you are so sure you can integrate it, post the solution here. If you notice, so far there have been 132 replies to this post and no person has found a solution yet. I hope you have enough dignity to stop this discussion immeditately or post the solution.

i havent inspected this closely, but i dont think this can be done. sandeep is right. you wont be able to get rid of the nCr terms..

thank you!!!!!!!!!

Replace

are correct!

Let y be the distance of the lowest point of the needle to the nearest line above it. If the needle falls horizontally, then y will be the distance from the needle to the nearest line above it. So,

. Let

represent the angle between the needle and the positive direction of x-axis, so that

. Then the ordered pair

determines the position of the needle.

So the square:

forms the sample space. The quantity

will be the the vertical height of the needle. Now, we know that the needle will intersect any one of the lines if and only if

. So the event space(favourable) is given by the area of region that lies between the curves

and

. Thus the required probability is equal to :

In our question ,

and

Is the answer

?

happy birthday sir!!!

Of course effective resistance cant be negative. So we reject the negative root.

Hence X =

Let

.

This problem can done by using the method of symmetry.

(If A and B are connected to battery and AB or it's perpendicular bisector is a line of symmetry, then all points lying on the perpendiculars drawn to AB are at same potential).

In the question, the perpendicular to AB is a line of symmetry. Hence the current through AF is equal to the current throguh FB which means no corrent flows from F to C or D. This implies that even if we break the connection at F, the currents in the circuit will not be affected. : Note that I have drawn the circuit only for the topmost layer of triangles.

Now calculate the resistance between CD.

Now simplifying the circuit, it consists of AB parallel to another branch (of resistance(

) .

Hence,

Now for the next layer of triangles, repeat the above procedure to see that

Now we see a pattern. For any particular layer is the resistance is

, the next layer the resistance is

.

So in effect the circuit is this:

Hence,

ohms

where

.